2007 AMC 12B — Problems & Online Timed Exam

Timed

1:15:00

1.

Isabella’s house has $3$ bedrooms. Each bedroom is $12$ feet long, $10$ feet wide, and $8$ feet high. Isabella must paint the walls of all the bedrooms. Doorways and windows, which will not be painted, occupy $60$ square feet in each bedroom. How many square feet of walls must be painted?

$678$

$768$

$786$

$867$

$876$

Answer: E

Concepts:perimeter area

Difficulty rating: 900

Small Hint:

The four walls of a room form a band of height $8$ wrapping around the floor

Big Hint:

The painted area per room is (floor perimeter) $\times$ (height) $-60$

Solution:

The perimeter of each bedroom floor is $2(12+10)=44$ feet.

So the wall area in one bedroom is $44\cdot8-60=352-60=292$ square feet. Across all three bedrooms, Isabella paints $3\cdot292=876$ square feet.

Thus, the correct answer is E.

2.

A college student drove his compact car $120$ miles home for the weekend and averaged $30$ miles per gallon. On the return trip the student drove his parents’ SUV and averaged only $20$ miles per gallon. What was the average gas mileage, in miles per gallon, for the round trip?

$22$

$24$

$25$

$26$

$28$

Answer: B

Concepts:distance rate and time harmonic mean

Difficulty rating: 990

Small Hint:

Average mileage is total miles divided by total gallons, not the average of the two rates

Big Hint:

The car uses $\frac{120}{30}$ gallons and the SUV uses $\frac{120}{20}$ gallons

Solution:

The trip home uses $\frac{120}{30}=4$ gallons, and the return trip uses $\frac{120}{20}=6$ gallons.

The round trip covers $240$ miles on $10$ gallons, so the average is $\dfrac{240}{10}=24.$

Thus, the correct answer is B.

3.

The point $O$ is the center of the circle circumscribed about $\triangle ABC,$ with $\angle BOC=120^\circ$ and $\angle AOB=140^\circ,$ as shown. What is the degree measure of $\angle ABC?$

$35$

$40$

$45$

$50$

$60$

Answer: D

Concepts:inscribed angle angle sum

Difficulty rating: 1190

Small Hint:

The three central angles at $O$ add up to $360^\circ$

Big Hint:

The inscribed angle $\angle ABC$ is half of the central angle $\angle AOC$

Solution:

The angles around $O$ sum to $360^\circ,$ so $\begin{aligned} \angle AOC&=360^\circ-140^\circ-120^\circ \\ &=100^\circ. \end{aligned}$

By the inscribed angle theorem, $\angle ABC$ subtends the same arc $AC$ as the central angle $\angle AOC,$ so $\angle ABC=\tfrac12\angle AOC=50^\circ.$

Thus, the correct answer is D.

4.

At Frank’s Fruit Market, $3$ bananas cost as much as $2$ apples, and $6$ apples cost as much as $4$ oranges. How many oranges cost as much as $18$ bananas?

$6$

$8$

$9$

$12$

$18$

Answer: B

Concepts:ratio and proportion unit conversion

Difficulty rating: 1010

Small Hint:

First convert $18$ bananas into an equivalent number of apples

Big Hint:

Then convert those apples into oranges using $6$ apples $=4$ oranges

Solution:

Since $3$ bananas cost as much as $2$ apples, $18$ bananas cost as much as $12$ apples.

Since $6$ apples cost as much as $4$ oranges, $12$ apples cost as much as $8$ oranges. Therefore $18$ bananas cost as much as $8$ oranges.

Thus, the correct answer is B.

5.

The $2007$ AMC $12$ contests will be scored by awarding $6$ points for each correct response, $0$ points for each incorrect response, and $1.5$ points for each problem left unanswered. After looking over the $25$ problems, Sarah has decided to attempt the first $22$ and leave the last $3$ unanswered. How many of the first $22$ problems must she solve correctly in order to score at least $100$ points?

$13$

$14$

$15$

$16$

$17$

Answer: D

Concepts:inequality

Difficulty rating: 1120

Small Hint:

The $3$ unanswered problems already contribute $3\cdot1.5$ points

Big Hint:

Solve $6x\ge100-4.5$ and round the number correct $x$ up to a whole number

Solution:

The three unanswered problems give $3\cdot1.5=4.5$ points.

So Sarah needs at least $100-4.5=95.5$ points from correct answers. Since $15\lt\dfrac{95.5}{6}\lt16,$ she must solve at least $16$ problems correctly, which would give her $16\cdot6+4.5=100.5$ points.

Thus, the correct answer is D.

6.

Triangle $ABC$ has side lengths $AB=5,$ $BC=6,$ and $AC=7.$ Two bugs start simultaneously from $A$ and crawl along the sides of the triangle in opposite directions at the same speed. They meet at point $D.$ What is $BD?$

$1$

$2$

$3$

$4$

$5$

Answer: D

Concepts:relative speed perimeter

Difficulty rating: 1080

Small Hint:

Together the two bugs cover the whole perimeter, so each crawls half of it

Big Hint:

One bug travels $A\to B\to D,$ a distance of $AB+BD$

Solution:

The perimeter is $5+6+7=18,$ so each bug crawls $9$ before they meet.

The bug going $A\to B\to C$ reaches $D$ on side $BC,$ having traveled $AB+BD=9.$ Since $AB=5,$ we get $BD=4.$

Thus, the correct answer is D.

7.

All sides of the convex pentagon $ABCDE$ are of equal length, and $\angle A=\angle B=90^\circ.$ What is the degree measure of $\angle E?$

$90$

$108$

$120$

$144$

$150$

Answer: E

Concepts:equilateral triangle square (geometry)angle sum

Difficulty rating: 1330

Small Hint:

The sides $AB,BC,EA$ and the two right angles make $ABCE$ a square

Big Hint:

Then $\triangle CDE$ is equilateral, and $\angle E=\angle AEC+\angle CED$

Solution:

Because $AB=BC=EA$ and $\angle A=\angle B=90^\circ,$ quadrilateral $ABCE$ is a square, so $\angle AEC=90^\circ$ and $EC$ equals the common side length.

Then $CD=DE=EC,$ so $\triangle CDE$ is equilateral and $\angle CED=60^\circ.$ Therefore $\begin{aligned} \angle E&=\angle AEC+\angle CED \\ &=90^\circ+60^\circ \\ &=150^\circ. \end{aligned}$

Thus, the correct answer is E.

8.

Tom’s age is $T$ years, which is also the sum of the ages of his three children. His age $N$ years ago was twice the sum of their ages then. What is $\frac{T}{N}?$

$2$

$3$

$4$

$5$

$6$

Answer: D

Concepts:ages linear equation

Difficulty rating: 1290

Small Hint:

$N$ years ago Tom was $T-N;$ each child was $N$ years younger

Big Hint:

The children then totaled $T-3N;$ set $T-N=2(T-3N)$

Solution:

Tom’s age $N$ years ago was $T-N.$ His three children were each $N$ years younger, so their ages then totaled $T-3N.$

The condition gives $T-N=2(T-3N),$ so $5N=T$ and $\frac{T}{N}=5.$

Thus, the correct answer is D.

9.

A function $f$ has the property that $f(3x-1)=x^2+x+1$ for all real numbers $x.$ What is $f(5)?$

$7$

$13$

$31$

$111$

$211$

Answer: A

Concepts:function substitution

Difficulty rating: 1200

Small Hint:

Pick $x$ so that $3x-1=5$

Big Hint:

With that $x,$ compute $x^2+x+1$

Solution:

Setting $3x-1=5$ gives $x=2.$

Then $f(5)=2^2+2+1=7.$

Thus, the correct answer is A.

10.

Some boys and girls are having a car wash to raise money for a class trip to China. Initially $40\%$ of the group are girls. Shortly thereafter two girls leave and two boys arrive, and then $30\%$ of the group are girls. How many girls were initially in the group?

$4$

$6$

$8$

$10$

$12$

Answer: C

Concepts:percentage invariant

Difficulty rating: 1330

Small Hint:

Two leave and two arrive, so the group size stays the same

Big Hint:

The two departing girls represent $40\%-30\%=10\%$ of the group

Solution:

Since two girls leave while two boys arrive, the total group size is unchanged. The drop from $40\%$ to $30\%$ girls corresponds to the two girls who left.

So those two girls are $10\%$ of the group, meaning the group has $20$ people. The initial number of girls was $40\%$ of $20,$ or $8.$

Thus, the correct answer is C.

11.

The angles of quadrilateral $ABCD$ satisfy $\angle A=2\angle B=3\angle C=4\angle D.$ What is the degree measure of $\angle A,$ rounded to the nearest whole number?

$125$

$144$

$153$

$173$

$180$

Answer: D

Concepts:angle sum ratio and proportion

Difficulty rating: 1270

Small Hint:

Let $x=\angle A;$ then $\angle B=\tfrac x2,$ $\angle C=\tfrac x3,$ $\angle D=\tfrac x4$

Big Hint:

The four angles of a quadrilateral sum to $360^\circ$

Solution:

Let $x=\angle A.$ Then $\angle B=\tfrac x2,$ $\angle C=\tfrac x3,$ and $\angle D=\tfrac x4.$

The angle sum gives $x+\dfrac x2+\dfrac x3+\dfrac x4=\dfrac{25x}{12}=360,$ so $x=\dfrac{12\cdot360}{25}=172.8\approx173.$

Thus, the correct answer is D.

12.

A teacher gave a test to a class in which $10\%$ of the students are juniors and $90\%$ are seniors. The average score on the test was $84.$ The juniors all received the same score, and the average score of the seniors was $83.$ What score did each of the juniors receive on the test?

$85$

$88$

$93$

$94$

$98$

Answer: C

Concepts:mean percentage

Difficulty rating: 1330

Small Hint:

Assume a convenient class of $10$ students: $1$ junior and $9$ seniors

Big Hint:

Total points must satisfy $10\cdot84=9\cdot83+s$

Solution:

Take a class of $10$ students, so there is $1$ junior and $9$ seniors.

The total of all scores is $10\cdot84=840,$ and the seniors contribute $9\cdot83=747.$ So the junior scored $840-747=93.$

Thus, the correct answer is C.

13.

A traffic light runs repeatedly through the following cycle: green for $30$ seconds, then yellow for $3$ seconds, and then red for $30$ seconds. Leah picks a random three-second time interval to watch the light. What is the probability that the color changes while she is watching?

$\dfrac{1}{63}$

$\dfrac{1}{21}$

$\dfrac{1}{10}$

$\dfrac{1}{7}$

$\dfrac{1}{3}$

Answer: D

Concepts:geometric probability

Difficulty rating: 1410

Small Hint:

One full cycle lasts $30+3+30=63$ seconds

Big Hint:

A change is seen only if her interval begins within $3$ seconds before one of the three color switches

Solution:

The cycle length is $30+3+30=63$ seconds, with three color changes per cycle.

Leah sees a change exactly when her three-second interval starts within the $3$ seconds before a switch. That gives $3\cdot3=9$ favorable seconds out of $63,$ a probability of $\dfrac{9}{63}=\dfrac17.$

Thus, the correct answer is D.

14.

Point $P$ is inside equilateral $\triangle ABC.$ Points $Q,$ $R,$ and $S$ are the feet of the perpendiculars from $P$ to $\overline{AB},$ $\overline{BC},$ and $\overline{CA},$ respectively. Given that $PQ=1,$ $PR=2,$ and $PS=3,$ what is $AB?$

$4$

$3\sqrt{3}$

$6$

$4\sqrt{3}$

$9$

Answer: D

Concepts:area decomposition equilateral triangle triangle area

Difficulty rating: 1680

Small Hint:

Split $\triangle ABC$ into $\triangle PAB,$ $\triangle PBC,$ $\triangle PCA,$ using the perpendiculars as heights

Big Hint:

Their areas sum to $\tfrac12 s(1+2+3),$ which also equals $\tfrac{\sqrt3}{4}s^2$

Solution:

Let $s=AB.$ Joining $P$ to the vertices splits the triangle into $\triangle PAB,$ $\triangle PBC,$ and $\triangle PCA,$ with areas $\tfrac{s}{2},$ $s,$ and $\tfrac{3s}{2}.$

Their total is $3s,$ which must equal the area $\tfrac{\sqrt3}{4}s^2$ of the equilateral triangle. So $3s=\dfrac{\sqrt3}{4}s^2,$ giving $s=\dfrac{12}{\sqrt3}=4\sqrt3.$

Thus, the correct answer is D.

15.

The geometric series $a+ar+ar^2+\cdots$ has a sum of $7,$ and the terms involving odd powers of $r$ have a sum of $3.$ What is $a+r?$

$\dfrac{4}{3}$

$\dfrac{12}{7}$

$\dfrac{3}{2}$

$\dfrac{7}{3}$

$\dfrac{5}{2}$

Answer: E

Concepts:geometric sequence

Difficulty rating: 1580

Small Hint:

The even-power terms sum to $7-3=4$

Big Hint:

The odd-power series is $r$ times the even-power series, so $3=r\cdot4$

Solution:

The odd-power terms are $ar+ar^3+\cdots$ $=r(a+ar^2+\cdots),$ that is, $r$ times the even-power terms. The even-power terms sum to $7-3=4.$

So $3=4r,$ giving $r=\tfrac34.$ Then $a=7(1-r)=\tfrac74,$ and $a+r=\dfrac74+\dfrac34=\dfrac52.$

Thus, the correct answer is E.

16.

Each face of a regular tetrahedron is painted either red, white, or blue. Two colorings are considered indistinguishable if two congruent tetrahedra with those colorings can be rotated so that their appearances are identical. How many distinguishable colorings are possible?

$15$

$18$

$27$

$54$

$81$

Answer: A

Concepts:Burnside’s Lemma symmetry

Difficulty rating: 2000

Small Hint:

The tetrahedron has $12$ rotational symmetries; apply Burnside’s lemma

Big Hint:

Count colorings fixed by the identity, the $8$ vertex rotations, and the $3$ edge rotations

Solution:

The rotation group of the tetrahedron has $12$ elements: the identity, $8$ rotations of order $3$ about a vertex-face axis, and $3$ rotations of order $2$ about an edge-midpoint axis.

The identity fixes all $3^4=81$ colorings. Each vertex rotation fixes one face and cycles the other three, so it fixes $3^2=9$ colorings; likewise each edge rotation swaps two pairs of faces and fixes $3^2=9.$

By Burnside’s lemma the number of distinguishable colorings is $\dfrac{81+8\cdot9+3\cdot9}{12}=\dfrac{180}{12}=15.$

Thus, the correct answer is A.

17.

If $a$ is a nonzero integer and $b$ is a positive number such that $ab^2=\log_{10}b,$ what is the median of the set $\{0,1,a,b,\frac{1}{b}\}?$

$0$

$1$

$a$

$b$

$\dfrac{1}{b}$

Answer: D

Concepts:logarithm inequality median (data)

Difficulty rating: 2060

Small Hint:

Since $b\lt10^b$ for all $b\gt0,$ we have $\log_{10}b\lt b$

Big Hint:

Rule out $b\ge1$ by checking whether $a=\dfrac{\log_{10}b}{b^2}$ can be a nonzero integer

Solution:

For $b\gt0,$ we have $b\lt10^b,$ hence $\log_{10}b\lt b.$ If $b=1,$ then $a=0.$ If $b\gt1,$ then $0\lt\dfrac{\log_{10}b}{b^2}\lt1,$ so $a$ could not be an integer.

Hence $0\lt b\lt1,$ so $\log_{10}b\lt0$ and $a=\dfrac{\log_{10}b}{b^2}\lt0.$ Thus $a\lt0\lt b\lt1\lt\dfrac1b,$ and the middle value of the sorted set is $b.$

Thus, the correct answer is D.

18.

Let $a,$ $b,$ and $c$ be digits with $a\ne0.$ The three-digit integer $\overline{abc}$ lies one third of the way from the square of a positive integer to the square of the next larger integer. The integer $\overline{acb}$ lies two thirds of the way between the same two squares. What is $a+b+c?$

$10$

$13$

$16$

$18$

$21$

Answer: C

Concepts:perfect square place value divisibility

Difficulty rating: 1930

Small Hint:

Let the two squares be $N^2$ and $(N+1)^2;$ their difference is $2N+1$

Big Hint:

Subtracting the two positions gives $\overline{acb}-\overline{abc}=9(c-b)=\tfrac{2N+1}{3}$

Solution:

Let the smaller square be $N^2,$ so the larger is $(N+1)^2$ and the gap is $2N+1.$ Then $\overline{abc}=N^2+\dfrac{2N+1}{3},$ $\overline{acb}=N^2+\dfrac{2(2N+1)}{3}.$

Subtracting, $\overline{acb}-\overline{abc}=9(c-b)$ $=\dfrac{2N+1}{3},$ so $27(c-b)=2N+1.$ Since $\overline{acb}$ is farther along the interval, $c-b$ is positive; and because the right side is odd, $c-b$ is odd. If $c-b\ge3,$ then $N\ge40$ and $N^2$ is not three digits.

So $c-b=1,$ giving $N=13.$ The points one third and two thirds of the way from $13^2=169$ to $14^2=196$ are $178$ and $187,$ so $a+b+c=1+7+8=16.$

Thus, the correct answer is C.

19.

Rhombus $ABCD,$ with side length $6,$ is rolled to form a cylinder of volume $6$ by taping $\overline{AB}$ to $\overline{DC}.$ What is $\sin(\angle ABC)?$

$\dfrac{\pi}{9}$

$\dfrac{1}{2}$

$\dfrac{\pi}{6}$

$\dfrac{\pi}{4}$

$\dfrac{\sqrt{3}}{2}$

Answer: A

Concepts:cylinder rhombus trigonometry

Difficulty rating: 1830

Small Hint:

Taping $\overline{AB}$ to $\overline{DC}$ makes a side of length $6$ the base circumference, so the radius is $\frac{3}{\pi}$

Big Hint:

The cylinder’s height is the rhombus altitude $6\sin\theta;$ set the volume equal to $6$

Solution:

Let $\theta=\angle ABC.$ The base circle has circumference $6,$ so its radius is $\dfrac{6}{2\pi}=\dfrac3\pi.$ The height of the cylinder is the rhombus altitude $6\sin\theta.$

The volume is $\pi\left(\dfrac3\pi\right)^2(6\sin\theta)=\dfrac{54}{\pi}\sin\theta=6,$ so $\sin\theta=\dfrac{\pi}{9}.$

Thus, the correct answer is A.

20.

The parallelogram bounded by the lines $y=ax+c,$ $y=ax+d,$ $y=bx+c,$ and $y=bx+d$ has area $18.$ The parallelogram bounded by the lines $y=ax+c,$ $y=ax-d,$ $y=bx+c,$ and $y=bx-d$ has area $72.$ Given that $a,$ $b,$ $c,$ and $d$ are positive integers, what is the smallest possible value of $a+b+c+d?$

$13$

$14$

$15$

$16$

$17$

Answer: D

Concepts:coordinate geometry parallelogram optimization

Difficulty rating: 2040

Small Hint:

The first parallelogram has area $\dfrac{(c-d)^2}{|b-a|};$ the second has area $\dfrac{(c+d)^2}{|b-a|}$

Big Hint:

Subtracting gives $4cd=54|b-a|;$ then minimize with small positive integers

Solution:

Two vertices of the first parallelogram lie at $(0,c)$ and $(0,d),$ and the other two have $x$ -coordinates $\pm\dfrac{c-d}{b-a}.$ Its area works out to $\dfrac{(c-d)^2}{|b-a|}=18.$ The same computation for the second gives $\dfrac{(c+d)^2}{|b-a|}=72.$

So $(c-d)^2=18|b-a|$ and $(c+d)^2=72|b-a|.$ Subtracting, $4cd=54|b-a|,$ i.e. $2cd=27|b-a|.$

Thus $|b-a|$ is even, so $a+b$ is smallest with $\{a,b\}=\{1,3\};$ and $cd$ is a multiple of $27,$ so $c+d$ is smallest with $\{c,d\}=\{3,9\}.$ These satisfy all conditions, giving $a+b+c+d=1+3+3+9$ $=16.$

Thus, the correct answer is D.

21.

The first $2007$ positive integers are each written in base $3.$ How many of these base- $3$ representations are palindromes? (A palindrome is a number that reads the same forward and backward.)

$100$

$101$

$102$

$103$

$104$

Answer: A

Concepts:number base palindrome casework

Difficulty rating: 2100

Small Hint:

Count base- $3$ palindromes by their number of digits; note $3^6=729\lt2007\lt2187=3^7$

Big Hint:

Total all palindromes with up to $7$ digits, then remove the $7$ -digit ones exceeding $2007=2202100_3$

Solution:

A palindrome is fixed by its first half. Counting base- $3$ palindromes by length gives $2$ of length $1$ or $2,$ $6$ of length $3$ or $4,$ $18$ of length $5$ or $6,$ and $54$ of length $7.$

That totals $2+2+6+6+18+18$ $+54=106$ palindromes with at most $7$ digits. Since $2007=2202100_3,$ the $7$ -digit palindromes larger than it are $2210122,$ $2211122,$ $2212122,$ $2220222,$ $2221222,$ and $2222222,$ which is $6$ of them.

Therefore the count is $106-6=100.$

Thus, the correct answer is A.

22.

Two particles move along the edges of equilateral $\triangle ABC$ in the direction $A\to B\to C\to A,$ starting simultaneously and moving at the same speed. One starts at $A,$ and the other starts at the midpoint of $\overline{BC}.$ The midpoint of the line segment joining the two particles traces out a path that encloses a region $R.$ What is the ratio of the area of $R$ to the area of $\triangle ABC?$

$\dfrac{1}{16}$

$\dfrac{1}{12}$

$\dfrac{1}{9}$

$\dfrac{1}{6}$

$\dfrac{1}{4}$

Answer: A

Concepts:similarity area ratio centroid

Difficulty rating: 2220

Small Hint:

Because both particles move linearly, the midpoint also moves linearly between key instants; call the traced triangle $XYZ$

Big Hint:

That triangle is centered at the centroid $O;$ compare its circumradius $OZ$ to $OC$

Solution:

Let $D,E,F$ be the midpoints of $BC,CA,AB,$ respectively, and let $X,Y,Z$ be the midpoints of $AD,BE,CF,$ respectively. Track a third point halfway between the two particles. It is at $X$ when the particles are at $A,D;$ at $Z$ when they are at $F,C;$ and at $Y$ when they are at $B,E.$

Between these instants both particle positions vary linearly, so their midpoint traces the segments $XZ,ZY,YX.$ Thus the enclosed path is the equilateral triangle $XYZ,$ which by symmetry shares the center $O$ of $\triangle ABC.$ Because $Z$ is the midpoint of the median $CF,$ $\begin{aligned} OZ&=OC-ZC \\ &=\dfrac23 CF-\dfrac12 CF \\ &=\dfrac16 CF, \end{aligned}$ while $OC=\dfrac23 CF.$

So the ratio of circumradii is $\dfrac{OZ}{OC}=\dfrac14,$ and the area ratio is $\left(\dfrac14\right)^2=\dfrac{1}{16}.$

Thus, the correct answer is A.

23.

How many non-congruent right triangles with positive integer leg lengths have areas that are numerically equal to $3$ times their perimeters?

$6$

$7$

$8$

$10$

$12$

Answer: A

Concepts:right triangle Diophantine Equation Simon’s Favorite Factoring Trick

Difficulty rating: 2140

Small Hint:

With legs $a\le b,$ write $\tfrac12 ab=3\left(a+b+\sqrt{a^2+b^2}\right)$ and isolate the radical

Big Hint:

Squaring and simplifying yields $(a-12)(b-12)=72;$ check the factor-pair candidates in the original equation to remove extraneous ones

Solution:

Let the legs be $a\le b.$ The condition is $\tfrac12 ab=3\left(a+b+\sqrt{a^2+b^2}\right),$ so $ab-6a-6b=6\sqrt{a^2+b^2}.$

Squaring and simplifying gives $ab(ab-12a-12b+72)=0,$ hence $(a-12)(b-12)=72.$ The positive integer solutions are $(a,b)=(3,4),$ $(13,84),$ $(14,48),$ $(15,36),$ $(16,30),$ $(18,24),$ $(20,21).$

The pair $(3,4)$ is extraneous: its area is $6,$ while its perimeter is $12$ and three times that is $36.$ So exactly $6$ triangles work.

Thus, the correct answer is A.

24.

How many pairs of positive integers $(a,b)$ are there such that $\gcd(a,b)=1$ and $\dfrac{a}{b}+\dfrac{14b}{9a}$ is an integer?

$4$

$6$

$9$

$12$

infinitely many

Answer: A

Concepts:divisibility greatest common divisor casework

Difficulty rating: 2340

Small Hint:

Clearing $b$ shows $\dfrac{14b^2}{9a}$ is an integer, and since $\gcd(a,b)=1,$ $14$ is divisible by $a$

Big Hint:

Similarly $9$ is divisible by $b,$ so $a\in\{1,2,7,14\}$ and $b\in\{1,3,9\};$ test the coprime pairs

Solution:

Let $k$ be the integer value of the original expression. Multiplying by $b$ and subtracting $a$ gives $\dfrac{14b^2}{9a}=bk-a,$ an integer. Since $\gcd(a,b)=1,$ it follows that $14$ is divisible by $a.$ Multiplying instead by $9a$ and subtracting $14b$ gives $\dfrac{9a^2}{b}=9ak-14b,$ so $9$ is divisible by $b.$

Thus $a\in\{1,2,7,14\}$ and $b\in\{1,3,9\}.$ Checking the coprime candidates, the expression is an integer only for $(a,b)=(1,3),$ $(2,3),$ $(7,3),$ $(14,3).$

So there are $4$ such pairs.

Thus, the correct answer is A.

25.

Points $A,$ $B,$ $C,$ $D,$ and $E$ are located in $3$ -dimensional space with $AB=BC=CD$ $=DE=EA=2$ and $\angle ABC=\angle CDE$ $=\angle DEA=90^\circ.$ The plane of $\triangle ABC$ is parallel to $\overline{DE}.$ What is the area of $\triangle BDE?$

$\sqrt{2}$

$\sqrt{3}$

$2$

$\sqrt{5}$

$\sqrt{6}$

Answer: C

Concepts:3D geometry coordinate geometry triangle area

Difficulty rating: 2400

Small Hint:

Place $D=(-1,0,0),$ $E=(1,0,0),$ and put $\triangle ABC$ in a horizontal plane $z=k\gt0$

Big Hint:

The right angles at $D$ and $E$ put $A,C$ on radius- $2$ circles; use $AC=2\sqrt2$ to find $k$

Solution:

Set $D=(-1,0,0)$ and $E=(1,0,0),$ and let $\triangle ABC$ lie in the plane $z=k\gt0.$ Because $\angle CDE$ and $\angle DEA$ are right angles, $A$ and $C$ lie on radius- $2$ circles centered at $E$ and $D$ in the planes $x=1$ and $x=-1,$ so $A=(1,y_1,k),\ C=(-1,y_2,k)$ with $y_j=\pm\sqrt{4-k^2}.$

Since $\angle ABC=90^\circ,$ $AC=2\sqrt2,$ which forces $y_1=-y_2.$ Taking $y_1=1,$ $y_2=-1,$ gives $k=\sqrt3,$ so $A=(1,1,\sqrt3),$ $C=(-1,-1,\sqrt3),$ and $B$ is $(1,-1,\sqrt3)$ or $(-1,1,\sqrt3).$

In the first case $BE=2$ with $BE\perp DE;$ in the second $BD=2$ with $BD\perp DE.$ Either way $\triangle BDE$ has legs $2$ and $2,$ so its area is $\tfrac12(2)(2)=2.$

Thus, the correct answer is C.